Question:
From the set
\[
\{2,3,5,7\}
\]
two numbers are selected one after the other with replacement. If \(X\) is the random variable representing the absolute difference of the two numbers selected, then the mean of \(X\) is
From the set
\[
\{2,3,5,7\}
\]
two numbers are selected one after the other with replacement. If \(X\) is the random variable representing the absolute difference of the two numbers selected, then the mean of \(X\) is
Show Hint
For discrete random variables,
\[
E(X)=\sum xP(X=x).
\]
When selections are made with replacement, first count all ordered outcomes, construct the probability distribution, and then calculate the expectation.
Updated On: Jul 9, 2026
- \(\dfrac{15}{4}\)
- \(\dfrac{17}{8}\)
- \(\dfrac{18}{5}\)
- \(6\) \bigskip
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
The mean (expected value) of a random variable \(X\) is
\[
E(X)=\sum x\,P(X=x).
\]
Since the selections are made with replacement, the total number of ordered pairs is
\[
4\times 4=16.
\]
Step 1: List all possible values of \(X\). Let \[ X=|a-b|, \] where \[ a,b\in\{2,3,5,7\}. \] The \(16\) ordered pairs and corresponding values of \(X\) are: \[ \begin{array}{c|c} (a,b) & X=|a-b| \hline (2,2),(3,3),(5,5),(7,7) & 0 (2,3),(3,2) & 1 (3,5),(5,3) & 2 (5,7),(7,5) & 2 (2,5),(5,2) & 3 (3,7),(7,3) & 4 (2,7),(7,2) & 5 \end{array} \]
Step 2: Find the probability distribution of \(X\). \[ P(X=0)=\frac{4}{16}=\frac14, \] \[ P(X=1)=\frac{2}{16}=\frac18, \] \[ P(X=2)=\frac{4}{16}=\frac14, \] \[ P(X=3)=\frac{2}{16}=\frac18, \] \[ P(X=4)=\frac{2}{16}=\frac18, \] \[ P(X=5)=\frac{2}{16}=\frac18. \]
Step 3: Compute the mean \(E(X)\). \[ E(X) = 0\left(\frac14\right) + 1\left(\frac18\right) + 2\left(\frac14\right) + 3\left(\frac18\right) + 4\left(\frac18\right) + 5\left(\frac18\right). \] \[ = 0+\frac18+\frac12+\frac38+\frac48+\frac58. \] \[ = \frac{1+4+3+4+5}{8}. \] \[ = \frac{17}{8}. \]
Step 4: Write the final answer. \[ \boxed{\frac{17}{8}} \]
Step 1: List all possible values of \(X\). Let \[ X=|a-b|, \] where \[ a,b\in\{2,3,5,7\}. \] The \(16\) ordered pairs and corresponding values of \(X\) are: \[ \begin{array}{c|c} (a,b) & X=|a-b| \hline (2,2),(3,3),(5,5),(7,7) & 0 (2,3),(3,2) & 1 (3,5),(5,3) & 2 (5,7),(7,5) & 2 (2,5),(5,2) & 3 (3,7),(7,3) & 4 (2,7),(7,2) & 5 \end{array} \]
Step 2: Find the probability distribution of \(X\). \[ P(X=0)=\frac{4}{16}=\frac14, \] \[ P(X=1)=\frac{2}{16}=\frac18, \] \[ P(X=2)=\frac{4}{16}=\frac14, \] \[ P(X=3)=\frac{2}{16}=\frac18, \] \[ P(X=4)=\frac{2}{16}=\frac18, \] \[ P(X=5)=\frac{2}{16}=\frac18. \]
Step 3: Compute the mean \(E(X)\). \[ E(X) = 0\left(\frac14\right) + 1\left(\frac18\right) + 2\left(\frac14\right) + 3\left(\frac18\right) + 4\left(\frac18\right) + 5\left(\frac18\right). \] \[ = 0+\frac18+\frac12+\frac38+\frac48+\frac58. \] \[ = \frac{1+4+3+4+5}{8}. \] \[ = \frac{17}{8}. \]
Step 4: Write the final answer. \[ \boxed{\frac{17}{8}} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions